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Binomial expansion with negative and fractional powers

Expand binomials with negative and fractional indices using exact coefficients. Understand infinite series, signs, substitutions and truncation with a manual graph and worked practice.

Before you startIndex laws, factorials and positive-integer binomial expansion

01 / From a finite expansion to an infinite series

A few powers can approximate a function near zero.

For a nonnegative integer exponent, the binomial expansion terminates. For negative or non-integer exponents, the same coefficient pattern generally continues forever. Keeping only the first few terms gives a polynomial approximation.

The model lets you select a power and a truncation degree. The written formula and error remain visible while you change the graph. Degree 0 keeps only the constant; degree 3 keeps terms through x³.

Choose how many powers to keepExplore

Blue solid: (1 + x)ᵖ. Gold dashed: the selected polynomial. Both are shown only for −0.5 ≤ x ≤ 0.5, inside the guaranteed convergence interval.

P₃(x) = 1 − x + x² − x³.

At x = 0.25: exact 0.8; polynomial 0.796875; absolute error 0.003125.

Exponent p Highest power kept 3Compare at x 0.25

Watch successive polynomial approximations

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Build the coefficients

Reduce the exponent factor by one at each stage.

(1 + z)ᵖ = 1 + pz + [p(p − 1)/2!]z²
+ [p(p − 1)(p − 2)/3!]z³ + …

For any real p, this infinite-series identity is guaranteed for |z| < 1. The first coefficient is 1, then p, then p(p − 1)/2, then p(p − 1)(p − 2)/6. Factorials 2! = 2 and 3! = 6 belong in the denominator.

If p is a nonnegative integer, a numerator factor eventually becomes zero and all later coefficients vanish. The resulting finite polynomial identity holds for all real z. The p = 0 case is simply 1 on the expression’s domain.

Do not give an infinite expansion a fixed number of terms such as “p + 1 terms” when p is negative or fractional. Endpoint behaviour at z = ±1 requires a separate convergence check; this lesson uses the guaranteed open interval.

03 / Rewrite roots and reciprocals first

Separate the exponent from the expression being substituted.

Read these as powers of a bracketWorked example

1/(1 − 3x)² = (1 − 3x)⁻²

The exponent is p = −2, while the substituted quantity is z = −3x.

√(1 + 6x) = (1 + 6x)^(½)

Here p = ½ and z = 6x.

1/∛(1 + 3x) = (1 + 3x)^(−⅓)

The reciprocal makes the cube-root exponent negative.

Substitute the whole quantity for z. For z = −3x, z² = 9x² and z³ = −27x³. The multiplier must also be raised to the required power.

04 / A negative integer exponent

Keep the signs from the coefficient and the bracket separate.

Expand (1 − 3x)⁻² through x³Worked example

Coefficients for p = −2: 1, −2, 3, −4

For example p(p − 1)(p − 2)/6 = (−2)(−3)(−4)/6 = −4.

1 − 2(−3x) + 3(−3x)² − 4(−3x)³ + …

Substitute z = −3x into every power.

= 1 + 6x + 27x² + 108x³ + …

The two negative signs in each odd-power contribution cancel.

Guaranteed for |−3x| < 1, so |x| < ⅓

The exponent p does not set the validity interval; the substituted z does.

At x = 0 the result must be 1. This catches an incorrect constant term, but it does not verify the other coefficients.

05 / A fractional exponent

Use exact fractions until the coefficients are simplified.

Expand √(1 + 6x) through x³Worked example

For p = ½: coefficients are 1, ½, −⅛, 1/16

The factors p − 1 and p − 2 are −½ and −3/2.

1 + ½(6x) − ⅛(6x)² + (1/16)(6x)³ + …

Raise 6x to each power.

= 1 + 3x − (9/2)x² + (27/2)x³ + …

Simplify each exact coefficient.

Guaranteed for |6x| < 1, so |x| < ⅙

In this interval the square-root input is positive.

A negative coefficient does not mean the whole function is negative. Several terms combine to approximate its value.

06 / Fractional exponents with a minus sign

A minus inside the bracket changes the odd powers.

Expand (1 − 2x)^(⅔) through x³Worked example

Coefficients for p = ⅔: 1, ⅔, −1/9, 4/81

Compute the falling products before substituting z.

1 + (⅔)(−2x) − (1/9)(−2x)² + (4/81)(−2x)³ + …

Even and odd powers have different signs.

= 1 − (4/3)x − (4/9)x² − (32/81)x³ + …

All three displayed nonconstant terms are negative for positive x.

Guaranteed for |x| < ½

This follows from |−2x| < 1.

Writing the unsimplified substitution line is a useful safeguard: p − 2 is not p/2, and (−2x)³ is not −2x³.

07 / A requested coefficient or term

The coefficient does not include the power of x.

Coefficient of xʲ in (1 + bx)ᵖ:
bʲ × p(p − 1)…(p − j + 1)/j!, for j ≥ 1

Find the x⁴ term in (1 + 2x)⁻¹Worked example

The coefficient from p = −1 is (−1)(−2)(−3)(−4)/24 = 1

Four factors correspond to the fourth power.

Multiply by 2⁴ to get coefficient 16

The x⁴ term is 16x⁴, not merely 16.

The constant is j = 0, so the term in xʲ is the (j + 1)th term when no earlier coefficient vanishes. “First four nonzero terms” and “through x³” can differ after a substitution such as z = x².

08 / An approximation is not an exact equality

An ellipsis and an approximation sign do different jobs.

For 1/(1 − x), keep powers through x²Worked example

1/(1 − x) = 1 + x + x² + x³ + … for |x| < 1

The ellipsis represents all the continuing terms.

1/(1 − x) ≈ 1 + x + x² near x = 0

Stopping after x² usually loses a nonzero remainder.

At x = 0.2: polynomial 1.24; exact value 1.25

The absolute error is 0.01.

Expand (1 − 2x)³. Does the usual infinite-series restriction |2x| < 1 apply?

Hint

A nonnegative integer exponent makes the expansion terminate.

Worked solution

1 − 6x + 12x² − 8x³ exactly. This polynomial identity holds for every real x; the infinite-series restriction is unnecessary.

10 / Recap

Keep the exponent, substituted quantity and truncation separate.

Rewrite roots and reciprocals as powers.
Use falling products divided by factorials.
Raise the entire substituted quantity to each power.
State the condition |z| < 1 for the infinite-series method.
Distinguish a coefficient from a term and a truncation from an exact identity.
Finite nonnegative-integer expansions are a separate case.

Review the positive-integer binomial theorem →

Section 1 of 10 · From a finite expansion to an infinite series